Chapter 6, Problem 6G.11E

(a)

Interpretation Introduction

Interpretation:

The $pH$ of the resulting solution after addition of base and the change in $pH$ has to be calculated.

Concept Introduction:

Henderson – Hasselbalch equation:

The $pH$ of a buffer can be easily calculated by using Henderson – Hasselbalch equation as given below.

  $pH=p{K}_a+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}$

Explanation of Solution

Given that, the $100mL$ buffer contains $0.100MNaC{H}_{3}C{O}_2\left(aq\right)$ and $0.100MC{H}_{3}COOH\left(aq\right).$

By using Henderson – Hasselbalch equation, the initial $pH$ of the above buffer solution can be calculated.

  $\begin{array}{c}pH=p{K}_a+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}\\ \\ pH=-\log \left(K_a\right)+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}\\ \\ =-\log \left(1.8\times {10}^{-5}\right)+\log \frac{0.100M}{0.100M}\\ \\ =4.75\end{array}$

The addition of a base will react with the weak acid and thus decreases the concentration of weak acid and increases the concentration of conjugate base.

The proton transfer equilibrium is given below.

  $\begin{array}{l}C{H}_{3}COOH\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+C{H}_{3}C{O}_2{}^{-}\left(aq\right)\\ \\ K_a=\frac{\left[H_3{O}^{+}\right]\left[C{H}_{3}C{O}_2{}^{-}\right]}{\left[C{H}_{3}COOH\right]}=1.8\times {10}^{-5}\end{array}$

The reaction of added base with some of $C{H}_{3}COOH$ is given below.

  $C{H}_{3}COOH\left(aq\right)+O{H}^{-}\left(aq\right)\to C{H}_{3}C{O}_2{}^{-}\left(aq\right)+H_{2}O\left(l\right)$

The initial number of moles of acid can be calculated as shown below.

  $\begin{array}{c}n{\left(C{H}_{3}COOH\right)}_{initial}=0.100L\times 0.100mol.L^{-1}\\ \\ =0.01mol\end{array}$

The number of moles of added base can be calculated as shown below.

  $\begin{array}{c}n{\left(NaOH\right)}_{initial}=0.01L\times 0.950mol.L^{-1}\\ \\ =0.0095mol\end{array}$

The molar ratio of acid and added base in the above equation is $1:1.$  The number of reacted moles of acid after addition of $0.0095molofNaOH$ can be calculated as given below.

  $\begin{array}{c}n{\left(C{H}_{3}COOH\right)}_{reacts}=0.0095molO{H}^{-}\times \frac{1molC{H}_{3}COOH}{1molO{H}^{-}}\\ \\ =0.0095molC{H}_{3}COOH\end{array}$

Now, the remaining amount of acid can be estimated.

  $\begin{array}{c}{n}_{final}=n_{initial}-n_{reacts}\\ \\ =0.01-0.0095\\ \\ =0.0005mol\\ \\ \left[C{H}_{3}COOH\right]=\frac{Moles}{Volume}=\frac{0.0005mol}{\left(0.100+0.01\right)L}=4.54\times {10}^{-3}M\end{array}$

The new concentration of conjugate base can be calculated as shown below.

  $\begin{array}{c}n{\left(C{H}_{3}CO{O}^{-}\right)}_{initial}=0.100L\times 0.100mol.L^{-1}\\ \\ =0.01mol\\ \\ n{\left(C{H}_{3}CO{O}^{-}\right)}_{final}=0.01mol+0.0095mol\\ \\ =0.0195mol\\ \\ \left[C{H}_{3}CO{O}^{-}\right]=\frac{Moles}{Volume}=\frac{0.0195mol}{\left(0.100+0.01\right)L}=0.177M\end{array}$

Now, the $pH$ of the buffer solution can be calculated as given below.

  $\begin{array}{c}pH=p{K}_a+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}\\ \\ pH=-\log \left(K_a\right)+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}\\ \\ =-\log \left(1.8\times {10}^{-5}\right)+\log \frac{0.177M}{4.54\times {10}^{-3}M}\\ \\ =\text{4}\text{.75+1}\text{.59}\\ \\ \text{=}6.34\\ \\ ChangeinpH=Final-initial=6.34-4.75=1.59\end{array}$

Therefore, the $pH$ of the resulting solution and the change in $pH$ is $6.34and1.59$ respectively.

(b)

Interpretation Introduction

Interpretation:

The $pH$ of the resulting solution after addition of acid and the change in $pH$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given that, the $100mL$ buffer contains $0.100MNaC{H}_{3}C{O}_2\left(aq\right)$ and $0.100MC{H}_{3}COOH\left(aq\right).$

By using Henderson – Hasselbalch equation, the initial $pH$ of the above buffer solution can be calculated.

  $\begin{array}{c}pH=p{K}_a+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}\\ \\ pH=-\log \left(K_a\right)+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}\\ \\ =-\log \left(1.8\times {10}^{-5}\right)+\log \frac{0.100M}{0.100M}\\ \\ =4.75\end{array}$

The addition of a strong acid will react with the conjugate base and thus decreases the concentration of conjugate base and increases the concentration of weak acid.

The proton transfer equilibrium is given below.

  $\begin{array}{l}C{H}_{3}COOH\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+C{H}_{3}C{O}_2{}^{-}\left(aq\right)\\ \\ K_a=\frac{\left[H_3{O}^{+}\right]\left[C{H}_{3}C{O}_2{}^{-}\right]}{\left[C{H}_{3}COOH\right]}=1.8\times {10}^{-5}\end{array}$

The reaction of added acid with some of $C{H}_{3}CO{O}^{-}$ is given below.

  $C{H}_{3}CO{O}^{-}\left(aq\right)+H^{+}\left(aq\right)\to C{H}_{3}C{O}_{2}H\left(aq\right)$

The initial number of moles of acid can be calculated as shown below.

  $\begin{array}{c}n{\left(C{H}_{3}COOH\right)}_{initial}=0.100L\times 0.100mol.L^{-1}\\ \\ =0.01mol\end{array}$

The number of moles of added acid can be calculated as shown below.

  $\begin{array}{c}n{\left(HN{O}_3\right)}_{initial}=0.02L\times 0.10mol.L^{-1}\\ \\ =0.002mol\end{array}$

The new concentration of weak acid can be calculated as shown below.

  $\begin{array}{c}n{\left(C{H}_{3}COOH\right)}_{final}=0.01mol+0.002mol\\ \\ =0.012mol\\ \\ \left[C{H}_{3}COOH\right]=\frac{Moles}{Volume}=\frac{0.012mol}{\left(0.100+0.02\right)L}=0.1M\end{array}$

The molar ratio of conjugate base and added acid is $1:1.$  The number of reacted moles of conjugate base after addition of $0.002molofHN{O}_3$ can be calculated as given below.

  $\begin{array}{c}n{\left(C{H}_{3}CO{O}^{-}\right)}_{reacts}=0.002mol{H}^{+}\times \frac{1molC{H}_{3}CO{O}^{-}}{1mol{H}^{+}}\\ \\ =0.002molC{H}_{3}CO{O}^{-}\end{array}$

Now, the remaining amount of conjugate base can be estimated.

  $\begin{array}{c}n{\left(C{H}_{3}CO{O}^{-}\right)}_{initial}=0.100M\times 0.100L\\ \\ =0.01mol\\ \\ n_{final}=n_{initial}-n_{reacts}\\ \\ =0.01-0.002\\ \\ =0.008mol\\ \\ \left[C{H}_{3}CO{O}^{-}\right]=\frac{Moles}{Volume}=\frac{0.008mol}{\left(0.100+0.02\right)L}=0.066M\end{array}$

Now, the $pH$ of the buffer solution can be calculated as given below.

  $\begin{array}{c}pH=p{K}_a+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}\\ \\ pH=-\log \left(K_a\right)+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}\\ \\ =-\log \left(1.8\times {10}^{-5}\right)+\log \frac{0.066M}{0.1M}\\ \\ \text{=}4.58\\ \\ ChangeinpH=Final-initial=4.58-4.75=-0.17\left(adecreaseof0.17\right)\end{array}$

Therefore, the $pH$ of the resulting solution and the change in $pH$ is $4.58and-0.17$ respectively.